If we assume the inner product is linear in the second argument, the polarization identity reads(adsbygoogle = window.adsbygoogle || []).push({});

[tex]

(x,y) = \frac 14 \| x + y \|^2 - \frac 14 \| x - y \|^2 - \frac i4 \|x + iy\|^2 + \frac i4 \| x - iy \|^2.

[/tex]

But there is another identity that I've seen referred to in some texts as the "polarization identity", and it is as follows: If T is a linear operator on the Hilbert space, we have

[tex]

(x,Ty) = \frac 14 (x+y,T(x+y)) - \frac 14 (x-y,T(x-y)) - \frac i4 (x+iy,T(x+iy)) + \frac i4 (x-iy,T(x-iy)).

[/tex]

I've expanded this out and checked to make sure it's correct, and it is...but is there a quick and painless way to prove this by starting from the first polarization identity above? If so, I really don't see it...my professor seemed to indicate to us that if we defined a kind of ersatz "inner product" on H by letting [itex](x,y)_1 = (x,Ty)[/itex], then the result would follow if we wrote out the first polarization identity with the norm [itex]\| \cdot \|_1[/itex] instead. But is this really kosher?

Also, just out of curiosity...why is it appropriate to call this thing a "polarization" identity?

**Physics Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# The polarization identity in Hilbert space

Loading...

Similar Threads for polarization identity Hilbert |
---|

A How to determine constant to be in Hilbert space |

I Polarization Formulae for Inner-Product Spaces ... |

Insights Hilbert Spaces And Their Relatives - Part II - Comments |

I Hilbert Space Example |

I Impossible to lift the identity map on the circle |

**Physics Forums | Science Articles, Homework Help, Discussion**